Ehrling-type Hölder and derivative interpolation with an epsilon loss

Reduction to $B_1$. Put $x_0$ for the centre of $B_R$, $v(z):=u(x_0+Rz)$ on $B_1$ and $g(z):=f(x_0+Rz)$. By the scaling identities of [F1] the three claims for $(u,f,R)$ are equivalent to the same claims for $(v,g,1)$: the factors $R^j$, $R^{k+\alpha }$, $R^{\alpha }$, $R^{-k}$ and $R^{\gamma }$ reproduce exactly the displayed powers. It therefore suffices to prove all three statements for $R=1$ with constants independent of the function; this is assumed from now on.

Part (iii). Let $0<\alpha <\beta \le 1$, $f\in C^{0,\beta }(B_1)$, $M:={\sup }_{B_1}\mathrm{\mid }f\mathrm{\mid }$ and $F:=[f{]}_{0,\beta ;B_1}$. If $M=0$ or $F=0$ the claim is immediate (for $F=0$, use that $f$ is constant, so $[f{]}_{0,\alpha }=0$; for $M=0$ the function vanishes), so assume $M,F>0$. If $\epsilon \ge 2^{\beta -\alpha }$, then for all $x\mathrm{\ne }y$ in $B_1$, $\mathrm{\mid }x-y\mathrm{\mid }\le 2$ and $[f{]}_{0,\alpha ;B_1}\le F\cdot 2^{\beta -\alpha }\le \epsilon F$, which is stronger than the claim; hence assume $\epsilon <2^{\beta -\alpha }$ and put $h_0:={\epsilon }^{1\mathrm{/}(\beta -\alpha )}\in (0,2)$. For a pair with $\mathrm{\mid }x-y\mathrm{\mid }\ge h_0$ use the trivial bound $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\le 2M$, and for a pair with $\mathrm{\mid }x-y\mathrm{\mid }<h_0$ use the $\beta$-Hölder bound: $\frac{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }}{\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }}\le \max \{2M{h}_0^{-\alpha },F{h}_0^{\beta -\alpha }\}=\max \{2M{\epsilon }^{-\alpha \mathrm{/}(\beta -\alpha )},F\epsilon \}\le F\epsilon +2M{\epsilon }^{-\alpha \mathrm{/}(\beta -\alpha )}\mathrm{.}$ This is the claim for $R=1$ with $C_3=2$.

Interior-point difference estimates, including points near the boundary. Write $M_j:={\max }_{\mathrm{\mid }\gamma \mathrm{\mid }=j}{\sup }_{B_1}\mathrm{\mid }{D}^{\gamma }u\mathrm{\mid }$, $M_0={\sup }_{B_1}\mathrm{\mid }u\mathrm{\mid }$, and $H:={\max }_{\mathrm{\mid }\gamma \mathrm{\mid }=k}[D^{\gamma }u{]}_{0,\alpha ;B_1}$, and set ${\rho }_{\ast }:=1\mathrm{/}(8nk)$. For each $x\in B_1$ choose an invertible frame $V_x$ as follows. If $\mathrm{\mid }x\mathrm{\mid }\le 1\mathrm{/}2$, take $V_x=I$. If $\mathrm{\mid }x\mathrm{\mid }>1\mathrm{/}2$, put $\nu =x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, choose an orthonormal basis $e_1,\dots ,e_{n-1}$ of the tangent space ${\nu }^{\perp }$, and take the columns of $V_x$ to be $e_i-\nu$ for $1\le i<n$ and $-\nu$ for the last column (when $n=1$ there is just the column $-\nu$). These frames and their inverses have norms bounded by constants depending only on $n$. Define ${\tilde{u}}_x(z):=u(x+V_{x}z)$ wherever $x+V_{x}z\in B_1$. For any vector $\mathrm{\ell }\in [0,\mathrm{\infty }{)}^n$ with $m:={\sum }_i{\mathrm{\ell }}_i\le nk$ and any $0\le t\le 1$, the whole segment $x+t\rho V_x\mathrm{\ell }$ lies in $B_1$ whenever $0<\rho \le {\rho }_{\ast }$. In the inner case, $\mathrm{\mid }x+t\rho V_x\mathrm{\ell }\mathrm{\mid }\le 1\mathrm{/}2+\rho m\le 5\mathrm{/}8$. In the outer case write $V_x\mathrm{\ell }=-m\nu +{\sum }_{i<n}{\mathrm{\ell }}_i{e}_i$; its tangential component has norm at most $m$, so $\mathrm{\mid }x+t\rho V_x\mathrm{\ell }{\mathrm{\mid }}^2\le r^2-2t\rho mr+2t^2{\rho }^2{m}^2\le r^2<1$, since $r=\mathrm{\mid }x\mathrm{\mid }\ge 1\mathrm{/}2$ and $t\rho m\le 1\mathrm{/}8\le r\mathrm{/}2$. Thus every sample point and Taylor segment below is contained in $B_1$. For $0\le d\le k-1$, choose the unique weights $a_0^{(d)},\dots ,a_{k-1}^{(d)}$ solving the Vandermonde system ${\sum }_{\mathrm{\ell }=0}^{k-1}{a}_{\mathrm{\ell }}^{(d)}{\mathrm{\ell }}^q=d!{\mathbf{1}}_{q=d}$ for $q=0,\dots ,k-1$. For a multi-index $\delta$ of order $j<k$, define the tensor stencil $Q_{\rho }^{\delta }{\tilde{u}}_x(0):={\rho }^{-j}{\sum }_{\mathrm{\ell }\in \{0,\dots ,k-1{\}}^n}\left({\prod }_{i=1}^n{a}_{{\mathrm{\ell }}_i}^{({\delta }_i)}\right){\tilde{u}}_x(\rho \mathrm{\ell })\mathrm{.}$ Taylor-expand ${\tilde{u}}_x$ at $0$ through degree $k-1$. The moment identities make this stencil equal to $D_z^{\delta }{\tilde{u}}_x(0)$ on every polynomial of total degree less than $k$. The integral remainder at each stencil point is bounded by $C_{n,k}{\rho }^k{M}_k$ using [F2] and the uniform frame bound. The weights and stencil are fixed by $n,k$, hence $\mathrm{\mid }{Q}_{\rho }^{\delta }{\tilde{u}}_x(0)-D_z^{\delta }{\tilde{u}}_x(0)\mathrm{\mid }\le C_{n,k}{\rho }^{k-j}{M}_k,\mathrm{\mid }{Q}_{\rho }^{\delta }{\tilde{u}}_x(0)\mathrm{\mid }\le C_{n,k}{\rho }^{-j}{M}_0\mathrm{.}$ For a multi-index $\delta$ of order $k$, instead use the ordinary iterated forward difference in the $z$-coordinates, ${\mathrm{\Delta }}_{\rho }^{\delta }{\tilde{u}}_x(0):={\sum }_{0\le \gamma \le \delta }(-1{)}^{k-\mathrm{\mid }\gamma \mathrm{\mid }}\left(\genfrac{}{}{0px}{}{\delta }{\gamma }\right){\tilde{u}}_x(\rho \gamma )$. Repeated use of the fundamental theorem of calculus gives ${\rho }^{-k}{\mathrm{\Delta }}_{\rho }^{\delta }{\tilde{u}}_x(0)$ as the average of $D_z^{\delta }{\tilde{u}}_x$ at points $\rho {\sum }_{q=1}^k{t}_q{e}_{i_q}$, where the list $i_1,\dots ,i_k$ contains ${\delta }_i$ copies of $i$ and $0\le t_q\le 1$. These segments lie in $B_1$ by the preceding geometry; the chain rule and the $\alpha$-Hölder seminorm of the order-$k$ derivatives therefore give $\mid {\rho }^{-k}{\mathrm{\Delta }}_{\rho }^{\delta }{\tilde{u}}_x(0)-D_z^{\delta }{\tilde{u}}_x(0)\mid \le C_{n,k}{\rho }^{\alpha }H,\mathrm{\mid }{\mathrm{\Delta }}_{\rho }^{\delta }{\tilde{u}}_x(0)\mathrm{\mid }\le 2^k{M}_0\mathrm{.}$ Finally, ${\mathrm{\partial }}_{x_a}={\sum }_i(V_x^{-1}{)}_{ia}{\mathrm{\partial }}_{z_i}$, so each canonical derivative of order $j$ is a uniformly bounded linear combination of frame derivatives of that order. We have proved, for every $x\in B_1$, $0<\rho \le {\rho }_{\ast }$, and canonical multi-index $\beta$, $\mathrm{\mid }{D}^{\beta }u(x)\mathrm{\mid }\le C_{n,k}({\rho }^{-j}{M}_0+{\rho }^{k-j}{M}_k)(j=\mathrm{\mid }\beta \mathrm{\mid }<k),\mathrm{\mid }{D}^{\beta }u(x)\mathrm{\mid }\le C_{n,k}({\rho }^{-k}{M}_0+{\rho }^{\alpha }H)(j=k)\mathrm{.}$ Taking suprema gives these same bounds for the full-ball quantities $M_j$; the estimates are valid up to points arbitrarily close to $\mathrm{\partial }{B}_1$.

Part (ii). By step 1.3, for every $x\in B_1$, $0<\rho \le {\rho }_{\ast }$, and $\mathrm{\mid }\beta \mathrm{\mid }=k$, $\mathrm{\mid }{D}^{\beta }u(x)\mathrm{\mid }\le C_{n,k}(M_0{\rho }^{-k}+{\rho }^{\alpha }H)\mathrm{.}$ If $M_0=0$, then $u=0$. If $H=0$ and $M_0>0$, use $\rho ={\rho }_{\ast }$ and take the supremum to obtain $M_k\le C(n,k)M_0$. Otherwise assume $M_0,H>0$ and put $s=(M_0\mathrm{/}H{)}^{1\mathrm{/}(k+\alpha )}$. If $s\ge {\rho }_{\ast }$, then $H\le {\rho }_{\ast }^{-(k+\alpha )}{M}_0$, and the estimate with $\rho ={\rho }_{\ast }$ gives $M_k\le C(n,k,\alpha )M_0$. If $s<{\rho }_{\ast }$, take $\rho =s$ to get $M_k\le C_{n,k,\alpha }{M}_0^{\alpha \mathrm{/}(k+\alpha )}{H}^{k\mathrm{/}(k+\alpha )}$. Young's inequality [F3], with $p=(k+\alpha )\mathrm{/}k$ and $q=(k+\alpha )\mathrm{/}\alpha$, then gives $M_k\le \epsilon H+C(n,k,\alpha ,\epsilon )M_0$. This proves (ii) for $R=1$.

Part (i). By part (ii), for every $\delta >0$ there is $C_{\delta }$ such that $M_k\le \delta H+C_{\delta }{M}_0$. Choose $\delta :=\epsilon \mathrm{/}(2k{C}_{n,k})$, increasing $C_{n,k}$ in step 1.3 if necessary so it is at least $1$. Fix any $\rho \in (0,{\rho }_{\ast }]$. For $0\le j<k$, the lower-order estimate of step 1.3 gives $M_j\le C_{n,k}{\rho }^{-j}{M}_0+C_{n,k}{\rho }^{k-j}{M}_k\le C(n,k,\alpha ,\epsilon )M_0+\frac{\epsilon }{2k}H,$ since ${\rho }_{\ast }<1$. Summing over the $k$ orders yields ${\sum }_{j=0}^{k-1}{M}_j\le C(n,k,\alpha ,\epsilon )M_0+\frac{\epsilon }{2}H\le C{M}_0+\epsilon \mathrm{\parallel }u{\mathrm{\parallel }}_{k,\alpha ;B_1}^{\ast }$, which is (i) for $R=1$.

Scaling back and conclusion. Undoing the change of variables of step 1.1 with the scaling identities of [F1] transforms (i), (ii) and (iii) for $R=1$ into the three displayed statements for general $R$, with the same constants: $R^j{M}_j(u)=M_j(v)$, $R^{k+\alpha }H(u)=H(v)$, $R^{\gamma }[f{]}_{0,\gamma ;B_R}=[g{]}_{0,\gamma ;B_1}$ and ${\sup }_{B_R}\mathrm{\mid }f\mathrm{\mid }={\sup }_{B_1}\mathrm{\mid }g\mathrm{\mid }$. The constants depend only on $n,k,\alpha ,\epsilon$ (and on $n,\alpha ,\beta ,\epsilon$ in (iii)), never on $u,f,R$ or the centre, and no choice principle is used.